ch04_06 - 4-51 SECTION 4.6 2 When solving differential equations of the form dy = f y dt for the unknown function y(t it is often convenient to make use

ch04_06 - 4-51 SECTION 4.6 2 When solving differential...

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4-51 SECTION 4.6 . . Integration by Substitution 393 2. When solving differential equations of the form dy dt = f ( y ) for the unknown function y ( t ), it is often convenient to make use of a potential function V ( y ). This is a function such that dV dy = f ( y ). For the function f ( y ) = y y 3 , find a potential function V ( y ). Find the locations of the local minima of V ( y ) and use a graph of V ( y ) to explain why this is called a “double- well” potential. Explain each step in the calculation dV dt = dV dy dy dt = − f ( y ) f ( y ) 0 . Since dV dt 0, does the function V increase or decrease as time goes on? Use your graph of V to predict the possible val- ues of lim t →∞ y ( t ). Thus, you can predict the limiting value of the solution of the differential equation without ever solving the equation itself. Use this technique to predict lim t →∞ y ( t ) if y = 2 2 y . 3. Let f n ( x ) = 2 n + 4 n 2 x 1 2 n x 0 2 n 4 n 2 x 0 x 1 2 n 0 otherwise for n = 1 , 2 , 3 , .... For an arbitrary n , sketch y = f n ( x ) and show that 1 1 f n ( x ) dx = 1. Compute lim n →∞ 1 1 f n ( x ) dx . For an arbitrary x = 0 in [ 2 , 2], compute lim n →∞ f n ( x ) and compute 1 1 lim n →∞ f n ( x ) dx . Is it always true that lim n →∞ 1 1 f n ( x ) dx = 1 1 lim n →∞ f n ( x ) dx ? 4.6 INTEGRATION BY SUBSTITUTION In this section, we expand our ability to compute antiderivatives by developing a useful technique called integration by substitution. This method gives us a process for helping to recognize a whole range of new antiderivatives. EXAMPLE 6.1 Finding an Antiderivative by Trial and Error Evaluate 2 xe x 2 dx . Solution We need to find a function F ( x ) for which F ( x ) = 2 xe x 2 . You might be tempted to guess that since x 2 is an antiderivative of 2 x , F ( x ) = x 2 e x 2 is an antiderivative of 2 xe x 2 . To see that this is incorrect, observe that, from the product rule, d dx ( x 2 e x 2 ) = 2 xe x 2 + x 2 e x 2 (2 x ) = 2 xe x 2 . So much for our guess. Before making another guess, look closely at the integrand. Notice that 2 x is the derivative of x 2 and x 2 already appears in the integrand, as the exponent of e x 2 . Further, by the chain rule, for F ( x ) = e x 2 , F ( x ) = e x 2 d dx ( x 2 ) = 2 xe x 2 , which is the integrand. To finish this example, recall that we need to add an arbitrary constant, to get 2 xe x 2 dx = e x 2 + c . More generally, recognize that when one factor in an integrand is the derivative of another part of the integrand, you may be looking at a chain rule derivative.
394 CHAPTER 4 . . Integration 4-52 Note that, in general, if F is any antiderivative of f , then from the chain rule, we have d dx [ F ( u )] = F ( u ) du dx = f ( u ) du dx . From this, we have that f ( u ) du dx dx = d dx [ F ( u )] dx = F ( u ) + c = f ( u ) du , (6.1) since F is an antiderivative of f . If you read the expressions on the far left and the far right sides of (6.1), this suggests that du = du dx dx .